A 2.60-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so

Question

A 2.60-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. A solid sphere of mass 0.25 kg is thrown horizontally with a speed

v1 = 15.5 m/s

to hit the rod at the point A one-fifth of the way up from the bottom of the rod. The sphere bounces back horizontally with a speed

v2 = 9.50 m/s,

while the rod swings to the right through an angle ? before swinging back toward its original position. What is the angular velocity of the rod immediately after the collision?
  
_____ rad/s

A 2.60-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. A solid sphere of mass 0.25 kg is thrown horizontally with a speed v1 = 15.5 m/s to hit the rod at the point A one-fifth of the way up from the bottom of the rod. The sphere bounces back horizontally with a speed v2 = 9.50 m/s, while the rod swings to the right through an angle ? before swinging back toward its original position. What is the angular velocity of the rod immediately after the collision?

Explanation / Answer

Note that the ball' lost KE must be the gained KE of the cylinder:          
Thus, for the ball,          

delta(KE) = 1/2mvf^2 - 1/2mvo^2 =    18.75   J  
          
Thus, this must be the KE of the cylinder right after contact.          

Thus, KE_cylinder =   18.75   J  

Note that for a cylinder rotating about its end,            
          
I = mL^2/3          
          
Thus, as m = 2.60 kg, L = 2.00 m,          
          
I =    3.466666667   kg*m^2  
          
Thus, as KE = 1/2 I w^2 ---> w = sqrt[2KE/I]:          
          
w =    3.29    rad/s   [ANSWER]

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.